## Calculus (3rd Edition)

$f'(x)= -x^{-2}+2x^{-3}=\frac{2}{x^{3}}-\frac{1}{x^{2}}$ Since the denominator can't be 0, $f'(0)$ doesn't exist. That is, 0 is a critical point of $f$. Another critical point is the solution of $f'(x)=0$ That is, $\frac{2}{x^{3}}-\frac{1}{x^{2}}=0$ $\implies \frac{2}{x^{3}}=\frac{1}{x^{2}}$ Or: $2=\frac{x^{3}}{x^{2}}=x$ Thus, the critical points are x=0, 2.